if alpha and beta are the zeroes of the polynomial 21y^2-y-2,find the Quadratic polynomial whose zeroes are 2alpha and 2 beta
Answers
Answer:
21y² - 2y - 4 = 0 hence proved
Step-by-step explanation:
21y² - y - 2 = 0
a= 21, b= -1 and c= -2
alpha × beta = c/a
= -2/21
multiplying 2 on both sides
2alpha×2beta= -4/21
alpha + beta = -b/a
= 1/21
multiplying 2 on both sides
2(alpha + beta)= 2/21
2alpha + 2beta = 2/21
new equation
a= 21 , b= -2 and c= -4
21y² - 2y - 4 = 0
Answer:
P(X) = 21x² - 26x + 8 .............
Step-by-step explanation:
Given polynomial is 21y² - y -2
Zeros of this polynomial is aplha and beta........
By quadratic formula........
-b +- root b² - 4(a)(c) /2a
a = 21
b = -1
c = -2
Therefore
1+- root 1-4(21)(-2)/42
1+-root 1+168 / 42
1+-root 169/42
1+-13/42
Case 1 ---------
1+13/42 = 14/42 = 1/3............ Alpha.
Case 2 ---------
1-13/42 = 12/42 = 2/7 ...........Betha.
Second polynomial zeros = 2alpha and 2betha .....
Therefore
zero 1 = 2*1/3 = 2/3.....
zero 2 = 2*2/7 =4/7......
Therefore (alpha) +(betha) = -b/a
2/3 + 4/7 = 26/21 ......
(alpha) *(betha) = 2/3 * 4/7 = 8/21.....
Therefore
a = 21
b = -26
c = 8
Therefore
P(X) = K[x² - (aplha +betha)x + (alpha*betha)......
= K[21x² - 26x + 8 ] ...........
Hope you understood
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